The total profit for the year is 19100. Of this B gets Tk. 9600 Therefore, A would get (19100 – 9600) = Tk 9500.

The partners split their profits in the ratio of their investments.

Therefore, the ratio of the investments of A : B = 9500 : 9600 = 95 : 96.

A invested Tk10000 initially for a period of 4 months. Then, he withdrew Tk2000.

Hence, his investment has reduced to Tk8000 (for the next 5 months).

Then he withdraws another Tk3000. Hence, his investment will stand reduced to Tk5000 during the last three months. So, the amount of money that he had invested in the company on a money-month basis will be = 4 * 10000 + 5 * 8000 + 3 * 5000 = 40000 + 40000 + 15000 = 95000

If A had 95000 money months invested in the company, B would have had 96,000 money months invested in the company (as the ratio of their investments is 95 : 96). If B had 96,000 money-months invested in the company, he has essentially invested 9600 

Go from the choices Choice (3) 30 and 20 are in the ratio of 6: 4 Five years ago their ages would be 25 and 15 which are in the ratio 5 : 3. Hence choice (3) is the right answer.

Amount of work done by 20 men = 24 women = 40 boys or amount of work done by 1 man = 1.2 women = 2 boys Let us therefore, find out the amount of men required, if only men were working on the task, to complete the new job under the new conditions. And then make adjustments for the women and boys who are already employed on the task. The man hours required to complete the new task = 4 times the man hours required to complete the old task. (As the new task is 4 times as big as the old task) So, the new task is 20 X 12 X 8 X 4 men hours task. Let 'n' be the number of men required to complete the new task. Equating the men hours of task, we get= 20 X 12 X 8 X 4  n = 128. The new task will require 128 men, if only men worked on the task to complete it.

However, the problem states that 6 women and 2 boys are working on the job. 6 women are the equivalent of 6/1.2 = 5 men and 2 boys = 1 man. i.e., the equivalen 

Let S be the selling price of 1 article. Therefore, the selling price of 100 articles = 100 S. --(1) The profit earned by selling these 100 articles = selling price of 75 articles = 75 S -- (2) We know that Selling Price (S.P.) = Cost Price (C.P) + Profit -- (3) Selling price of 100 articles = 100 S and Profit = 75 S from (1) and (2). Substituting this in eqn (3), we get 100 S = C.P + 75 S. Hence, C.P = 100 S - 75 S = 25 S. Profit % = 300%

The arithmetic mean of 5 observations is $7000. Therefore, the sum of these 5 observations is 5 * 7000 = $35,000. The median is $7000. Let us say their incentives in ascending order are a, b, c, d and e. So, c = 7000 and a + b + c + d + e = $35,000 The only mode is $12,000. So, the maximum number of observations among the 5 will be in $12,000. The only possibility is that both d and e got an incentive of $12,000 each. So, c + d + e = 7000 + 12000 + 12000 = $31,000 Therefore, a + b = 35,000 - 31,000 = 4000. a and b have to be two different values as the only mode is $12,000. So, a has to be $1000 and b has to be $3000. The difference between the highest and the lowest is therefore 12,000 - 1000 = $11,000

The new average weight of the group after including the football coach = 31 As the new average is 1kg more than the old average, old average without including the football coach = 30 kgs. The total weight of the 30 friends without including the football coach = 30 X 30 = 900. After including the football coach, the number people in the group increases to 31 and the average weight of the group increases by 1kg.

Therefore, the total weight of the group after including the weight of the football coach = 31 X 31 = 961 kgs. Therefore, the weight of the football coach = 961 - 900 = 61 kgs.

The new mean  =62+{(128+255+511+1023+x)−(28+x+42+78+104)}/5                          =62+{(1917+x)−(252+x)}/5